Chapter 3, Problem 80P

An open settling tank shown in the figure contains a liquid suspension. Determine the resultant force acting on the gate and its line of action if the liquid density is 850 kg/m³. The gate is parabolic as sketched, looking straight at the gate.

To determine

The Resultant force acting on gate and its line of action.

Answer to Problem 80P

The Resultant force acting on gate is $140.524\text{kN}$ and its line of action of the hydrostatic force of liquid from the bottom is at $1.63\text{m}$.

Explanation of Solution

Given:

Density of liquid is $850\text{kg}/{\text{m}}^{\text{3}}$, the gate is parabolic.

Draw the cross-sectional view of the gate.

  

Figure (1)

Write the expression for the curve.

  $x=\sqrt{\frac{y}{2}}$   ...... (I)

Here, horizontal axis is denoted by the x and vertical axis is denoted by the y.

Write the expression for the vertical distance between the centre of the elemental axis and the x axis.

  $\stackrel{-}{y}=y$   ...... (II)

Here, the vertical distance between the centre of the elemental axis and the x axis is $\stackrel{-}{y}$.

Write the expression for the area of the elemental strip.

  $dA=2xdy$   ...... (III)

Here, area is $dA$.

Write the expression for the centre of gravity.

  $CG=\frac{{\displaystyle \underset{0}{\overset{y_{\max }}{\int }}\stackrel{-}{y}dA}}{{\displaystyle \underset{0}{\overset{y_{\max }}{\int }}dA}}$   ...... (IV)

Here, centre of gravity is $CG$.

Substitute $3$ for $y_{\max }$, Equation (I), (II), and (III) in Equation (IV).

  $CG=\frac{{\displaystyle \underset{0}{\overset{3}{\int }}y\left(2xdy\right)}}{{\displaystyle \underset{0}{\overset{3}{\int }}2xdy}}$   ...... (V)

Draw a view of gate and liquid.

  

Figure (2)

Write the expression for the length of gate which is in contact with the liquid.

  $l=\frac{z}{\sin 60°}$...... (VI)

Here, length of the gate is $l$ and perpendicular height from free surface is $z$.

Write the expression for the centre of gravity of the gate from the free surface.

  $h_{CG}=I-CG$   ...... (VII)

Here, the mass moment of inertia is $I$ and the centre of gravity of the gate from the free surface is $h_{CG}$.

Write the expression for the vertical depth of the centre of gravity of gate from the free surface.

  $h^{\text{'}}=h_{CG}\sin \beta$   ...... (VIII)

Here, the vertical depth of the centre of gravity of gate from the free surface is $h^{\text{'}}$.

Write the expression for pressure acing on the gate.

  $P_a=\rho g{h}^{\text{'}}$   ...... (IX)

Here, pressure acing on the gate is $P_a$.

Write the expression for the area of gate resisting the pressure on the gate.

  $A_{gate}={\displaystyle \underset{0}{\overset{3}{\int }}2xdy}$   ...... (X)

Here, the area of gate resisting the pressure on the gate is $A_{gate}$.

Write the expression for the resultant force acting on the gate.

  $F_R=P_a{A}_{gate}$   ...... (XI)

Here, the resultant force acting on the gate is $F_R$.

Write the expression for the moment of inertia.

  $I={\displaystyle \underset{0}{\overset{3}{\int }}{y}^{2}dA}$   ...... (XII)

Write the expression for the action of resultant hydrostatic force.

  $y_p=CG+\frac{I}{CG\cdot A}$   ...... (XIII)

Here, the action of resultant hydrostatic force is $y_p$.

Write the expression for the centre of line of action from the free surface.

  $h_p=y_p\sin 60°$   ...... (XIV)

Here, the centre of line of action from the free surface is $h_p$.

Write the expression for the centre of location of the force from the bottom.

  $H=5\text{m}-h_p$   ...... (XV)

Here, the centre of location of the force from the bottom is $H$.

Calculation:

Substitute $\sqrt{\frac{y}{2}}$ for $x$ in Equation (V).

  $\begin{array}{c}CG=\frac{{\displaystyle \underset{0}{\overset{3}{\int }}y\left(2\sqrt{\frac{y}{2}}dy\right)}}{{\displaystyle \underset{0}{\overset{3}{\int }}2\sqrt{\frac{y}{2}}dy}}\\ =\frac{{\displaystyle \underset{0}{\overset{3}{\int }}\left(y\sqrt{y}\right)dy}}{{\displaystyle \underset{0}{\overset{3}{\int }}\left(\sqrt{y}\right)dy}}\\ ={\left[\frac{\left(\frac{y^{\frac{5}{2}}}{5/2}\right)}{\left(\frac{y^{3/2}}{3/2}\right)}\right]}_0^3\end{array}$

  $\begin{array}{c}={\left(\frac{3}{5}y\right)}_0^3\\ =\frac{9}{5}\text{m}\\ \text{=1}\text{.8}\text{m}\end{array}$

Substitute $5\text{m}$ for $z$ in Equation (VI).

  $\begin{array}{c}l=\frac{5\text{m}}{\sin 60°}\\ =\frac{5\text{m}}{\left(\frac{\sqrt{3}}{2}\right)}\\ =\frac{10}{\sqrt{3}}\text{m}\end{array}$

Substitute $\frac{10}{\sqrt{3}}\text{m}$ for $l$ and $\frac{9}{5}\text{m}$ for $CG$ in Equation (VII).

  $\begin{array}{c}{h}_{CG}=\left(\frac{10}{\sqrt{3}}\text{m}-\frac{9}{5}\text{m}\right)\\ =5.7735\text{m}-1.8\text{m}\\ \text{=3}\text{.9735}\text{m}\end{array}$

Substitute $\text{3}\text{.9735}\text{m}$ for $h_{CG}$ and $60°$ for $\beta$ in Equation (VIII).

  $\begin{array}{c}{h}^{\text{'}}=3.97\text{m}\times \sin 60°\\ =3.97\text{m}\times 0.866\\ =3.44\text{m}\end{array}$

Substitute $850\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $3.44\text{m}$ for $h^{\text{'}}$ in Equation (IX).

  $\begin{array}{c}{P}_a=850\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}\times 3.44\text{m}\\ =838.5\times 3.44\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\\ =28684.44\text{N}/{\text{m}}^2\end{array}$

Substitute $\sqrt{\frac{y}{2}}$ for $x$ in Equation (X).

  $\begin{array}{c}{A}_{gate}={\displaystyle \underset{0}{\overset{3}{\int }}2\sqrt{\frac{y}{2}}}dy\\ =\sqrt{2}{\displaystyle \underset{0}{\overset{3}{\int }}\sqrt{y}}dy\\ =\sqrt{2}{\left[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^3\\ =2\sqrt{6}{\text{m}}^2\end{array}$

Substitute $28684.44\text{N}/{\text{m}}^2$ for $P_a$ and $2\sqrt{6}{\text{m}}^2$ for $A_{gate}$ in Equation (XI).

  $\begin{array}{c}{F}_R=28684.44\text{N}/{\text{m}}^2\times 2\sqrt{6}{\text{m}}^2\\ =140524.483\text{N}\left(\frac{1\text{kN}}{1000\text{N}}\right)\\ =140.524\text{N}\end{array}$

Substitute $\sqrt{\frac{y}{2}}$ for $x$ in Equation (XII).

  $\begin{array}{c}I={\displaystyle \underset{0}{\overset{3}{\int }}2\left(\sqrt{\frac{y}{2}}\right)}{y}^{2}dy\\ =\sqrt{2}{\displaystyle \underset{0}{\overset{3}{\int }}{y}^{\frac{5}{2}}dy}\\ =\sqrt{2}{\left(\frac{y^{7/2}}{7/2}\right)}_0^3\\ =18.896{\text{m}}^4\end{array}$

Substitute $1.8\text{m}$ for $CG$, $18.896{\text{m}}^4$ for $I$ and $2\sqrt{6}{\text{m}}^2$ for $A_{gate}$ in Equation (XIII).

  $\begin{array}{c}{y}_p=1.8\text{m}+\frac{18.896{\text{m}}^4}{\left(1.8\text{m}\times 2\sqrt{6}{\text{m}}^2\right)}\\ =1.8\text{m}+2.1428\text{m}\\ =3.9\text{m}\end{array}$

Draw the diagram for different parameter.

  

Figure (3)

Substitute $3.9\text{m}$ for $y_p$ in Equation (XIV).

  $\begin{array}{c}{h}_p=3.9\text{m}\times \sin 60°\\ =3.9\text{m}\times 0.866\\ =3.37\text{m}\end{array}$

Substitute $3.37\text{m}$ for $h_p$ in Equation (XV).

  $\begin{array}{c}H=5\text{m}-3.37\text{m}\\ =1.63\text{m}\end{array}$

Conclusion:

Resultant force acting on gate is $140.524\text{kN}$ and its line of action of the hydrostatic force of liquid from the bottom is at $1.63\text{m}$.